Problem Statement
Find the coordinates of the foot of the perpendicular drawn from the point $P(1, 2, 3)$ to the line $\frac{x - 6}{3} = \frac{y - 7}{2} = \frac{z - 7}{-2}$. Also, find the length of the perpendicular.
Verified Solution & Marking Scheme
Express coordinates of general point on the line
Let $\frac{x - 6}{3} = \frac{y - 7}{2} = \frac{z - 7}{-2} = \lambda$.
A general point $Q$ on the line has coordinates:
$Q(3\lambda + 6, \, 2\lambda + 7, \, -2\lambda + 7)$
Form direction ratios of PQ
Point $P = (1, 2, 3)$. Direction ratios of vector $\vec{PQ}$ are:
$a = (3\lambda + 6 - 1) = 3\lambda + 5$
$b = (2\lambda + 7 - 2) = 2\lambda + 5$
$c = (-2\lambda + 7 - 3) = -2\lambda + 4$
Apply perpendicularity condition PQ ⟂ line
Direction ratios of the given line are $(3, 2, -2)$.
Since $PQ \perp$ line:
$3(3\lambda + 5) + 2(2\lambda + 5) + (-2)(-2\lambda + 4) = 0$
$9\lambda + 15 + 4\lambda + 10 + 4\lambda - 8 = 0$
$17\lambda + 17 = 0 \implies \lambda = -1$
Determine coordinates of Q and distance PQ
Substitute $\lambda = -1$ into $Q$:
$x = 3(-1) + 6 = 3, \quad y = 2(-1) + 7 = 5, \quad z = -2(-1) + 7 = 9$
Foot of perpendicular: $Q(3, 5, 9)$.
Length of perpendicular:
$|PQ| = \sqrt{(3 - 1)^2 + (5 - 2)^2 + (9 - 3)^2} = \sqrt{2^2 + 3^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7 \text{ units}$